Question

A ball is thrown towards a wall with an initial speed of 25.0 m/s, starting at 8.20 m above the ground. The height of the wall is 10.0 m and it is located 20.0 m from the launch point of the ball. You may assume projectile motion for the ball. (SEE DIAGRAM IN PICTURE BELOW)

a) When the ball passes directly above the wall, how high above the wall is it? What is the velocity of the ball when it passes directly above the wall? Give magnitude and direction for the velocity.

b.) How far from the wall does the ball land when it reaches the ground?

25.0 m/s l. A ball is thrown towards a wall with an initial speed of 25.0 m/s, starting at 40.0 8.20 m above the ground. The height of the wall is 10.0 m and it is located 8.20 m 10.0 m. 20.0 m from the launch point of the ball. See figure to the right. You may assume projectile motion for the ball 20.0 m. (a) When the ball passes directly above the wall, how high above the wall is it? What is the of the ball when it passes directly above the wall? Give magnitude and direction for the velocity (b) How far from the wall does the ball land when it reaches the ground?

Answer 1

Horizontal distance travelled to reach wall is x = 20 m

but x = u*cos(theta)*T

T is the time of flight

x = 25*cos(40)*T = 20

T = 1.04 sec

vertical distace travelle is y = (u*sin(40)*T)-(0.5*g*T^2) = (25*sin(40)*1.04)-(0.5*9.8*1.04^2) = 11.41 m

required height above the wall is 11.41-10 = 1.41 m

horizontal component of velocity is Vx = Vox = 25*cos(40) = 19.15 m/sec

vertical component of velocity is Vy = Voy- (g*T) = (25*sin(40))-(9.8*1.04) = 5.87 m/sec

V = sqrt(Vx^2+Vy^2) = sqrt(19.15^2+5.87^2) = 20.02 m/sec

direction is tan(theta) = Vy/Vx = 5.87/19.15

theta = tan^(-1)(5.87/19.15) = 17 degrees above the X-axis

b) actual range is R = u^2*sin(2*theta)/g = 25^2*sin(2*40)/9.8 = 62.8 m

then required horizontal distance is 62.8-20 = 42.8 m