An apple is thrown at an angle of 30° above the horizontal from the top of a building 20 m high. Its initial speed is 40 ms-1. Calculate

Question

An apple is thrown at an angle of 30° above the horizontal from the top of a building 20 m high. Its initial speed is 40 ms-1. Calculate
a)the time taken for the apple to strikes the ground
b) the distance from the foot of the building will it strikes the ground
c) the maximun height reached by the apple from the ground​

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Amity 4 years 2021-08-03T11:57:30+00:00 1 Answers 31 views 0

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  1. Sapo1
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    2021-08-03T11:58:44+00:00
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    Answer:

    (a) 4.83 seconds

    (b) 167.3m

    (c) 40m

    Explanation:

    This is a two-dimensional motion. Therefore, the components of the initial velocity – u_X and u_Y – in the x and y directions are given as follows:

    u_X = u cos θ                 ————–(*)

    u_Y = u sin θ                ————-(**)

    Where

    θ = angle of projection

    (a) To calculate the time taken for the apple to strike the ground.

    For simplicity, let’s first calculate the maximum height H reached by the apple.

    Using one of the equations of motion as follows, we can find H:

    v² = u² + 2as               —————(i)

    Where;

    v = velocity at maximum height = 0 [at maximum height, velocity is 0]

    u = initial vertical velocity of the apple = u_Y

    => u = u_Y = u sin θ

    => u_Y = 40 sin 30°

    => u_Y = 40 x 0.5

    => u_Y = 20 m/s

    a = acceleration due to gravity = -g [apple moves upwards against gravity]

    a = -10m/s²

    s = H = maximum height reached from the top of the building

    Substitute these values into equation (i) above to have;

    0² = u_Y² + 2aH

    0² = (20)² + 2(-10)H

    0 = 400 – 20H

    20H = 400

    H = 20m

    The total time taken to strike the ground is the sum of the time taken to reach maximum height and the time taken to strike the ground from maximum height.

    =>Calculate time t₁ to reach maximum height.

    Using one of the equations of motion, we can calculate t₁ as follows;

    v = u_Y  + at                —————(ii)

    Where;

    v = velocity at maximum height = 0

    u = initial vertical velocity of the apple = u_Y

    a = -g = -10m/s²           [acceleration due to gravity is negative since the apple is thrown upwards to reach maximum height]

    t = t₁ = time taken to reach maximum height.

    Equation (ii) then becomes;

    0 = 20  + (-10)t₁

    10t₁ = 20

    t₁ = 2 seconds

    =>Calculate time t₂ to strike the ground from maximum height.

    Now, using one of the equations of motion, we can calculate the time taken as follows;

    Δy = u_Y t + \frac{1}{2}at                —————(iii)

    Where;

    Δy = displacement from maximum height to the ground = maximum height from top of building + height of building = 20 + 20 = 40m

    a = g = 10m/s²            [acceleration due to gravity is positive since the apple is now coming downwards from maximum height]

    t = t₂    [time taken to strike the ground from maximum height]

    u_Y = initial vertical velocity from maximum height = 0

    u_Y = 0

    Equation (iii) then becomes

    40 = 0t + \frac{1}{2}(10)t₂²

    40 = 5t₂²            [divide through by 5]

    8 = t₂²

    t₂ = ±\sqrt{8}

    t₂ = ±2\sqrt{2}

    t₂ = +2\sqrt{2} or -2\sqrt{2}

    since time cannot be negative,

    t₂ = 2\sqrt{2} = 2.83 seconds

    Therefore, the time taken for the apple to strike the ground is;

    t₁ + t₂ = 2 +  2.83 = 4.83 seconds

    (b) The distance from the foot of the building where the apple will strike the ground

    Since this is the horizontal distance, we use the horizontal version of equation (iii) as follows;

    Δx = u_X t + \frac{1}{2}at       ———–(v)

    Where

    Δx = distance from the foot of the building to where the apple strikes the ground.

    u_X = initial horizontal velocity of the apple as expressed in equation (*)

    u_X = 40 cos 30

    u_X = 34.64 m/s

    t = time taken for the motion of the apple = 4.83 seconds [calculated above]

    a = acceleration due to gravity in the horizontal direction = 0. [For a projectile, there is no acceleration in the horizontal direction since velocity is constant]

    Substitute these values into equation (v) as follows;

    Δx = u_X t + \frac{1}{2}(0)t

    Δx = 34.64 x 4.83

    Δx = 167.3 m

    Therefore, the distance from the foot of the building is 167.3m

    (c) The maximum height reached by the apple from the ground

    This is the sum of the height reached from the top of the building (20m which has been calculated in (a) above) and the height of the building.

    = 20m + 20m = 40m

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