Question

A playground is on the flat roof of a city school, 6.5 m above the street below (see figure). The vertical wall of the building is h = 7.80 m high, forming a 1.3-m-high railing around the playground. A ball has fallen to the street below, and a passerby returns it by launching it at an angle of θ = 53.0° above the horizontal at a point d = 24.0 m from the base of the building wall. The ball takes 2.20 s to reach a point vertically above the wall. (a) Find the speed at which the ball was launched. (Give your answer to two decimal places to reduce rounding errors in later parts.) m/s (b) Find the vertical distance by which the ball clears the wall. m (c) Find the horizontal distance from the wall to the point on the roof where the ball lands.

Answer 1

Part A.

In horizontal motion range of projectile is given by:

R = V0x*t

V0x = Initial horizontal speed = V0*cos A

A = launching angle = 53.0 deg

at t = 2.20 sec, R = 24.0 m, So

24.0 = V0*cos 53.0 deg*2.20 sec

V0 = 24.0/(2.20*cos 53 deg)

V0 = 18.13 m/sec = Initial launching speed

Part B.

At this time height of ball will be:

Using 2nd kinematic equation:

h = V0y*t + (1/2)*ay*t^2

ay = Vertical acceleration = -9.81 m/sec^2

V0y = V0*sin A = 18.13*sin 53.0 deg = 14.48 m/sec

t = 2.20 sec

So,

h = 14.48*2.20 + (1/2)*(-9.81)*2.20^2

h = 8.12 m

Since given that wall's height = 7.80 m

So, distance by which ball clears the wall = 8.12 - 7.80 = 0.32 m

Part C.

Now when ball will land on roof, it's height from ground will be: h = 6.50 m (since given that playground is 6.50 above the street), So to find the time when ball lands on playground,

Again using 2nd kinematic equation:

h = V0y*t + (1/2)*ay*t^2

6.50 = 14.48*t + (1/2)*(-9.81)*t^2

4.905*t^2 - 14.48*t + 6.50 = 0

Solving above quadratic equation

t1 = [14.48 +/- sqrt (14.48^2 - 4*4.905*6.50)]/(2*4.905)

t1 = 2.40 sec

(we will use root given by +sign, since time given by -ve sign gives the time before ball reaches maximum height)

Now at this ball's horizontal distance will be:

R1 = V0x*t1

R1 = V0*cos A*t1

R1 = 18.13*2.40*cos 53 deg

R1 = 26.19 m = Horizontal distance of ball from launching point

Now this distance R1 is from the launching point, So distance of ball from wall will be:

d1 = R1 - R

d1 = 26.19 - 24.0 = 2.19 m

d1 = 2.19 m = distance from wall to the point on the roof where the ball lands

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