A home run is hit in such a way that the baseball just clears a wall 10.0 m high

Question

A home run is hit in such a way that the baseball just clears a wall 10.0 m high, located 114 m from home plate. The ball is hit at an angle of 33.0° to the horizontal, and air resistance is negligible. (Assume that the ball is hit at a height of 1.0 m above the ground.) (a) Find the initial speed of the ball. m/s (b) Find the time it takes the ball to reach the wal 12.13 Use the initial speed you found to determine how long it takes the baseball to reach the 114 m distance measured along the ground. s (c) Find the velocity components of the ball when it reaches the wal x-component y-component m/s m/s Find the speed of the ball when it reaches the wall. m/s

Explanation / Answer

a)
let vo is the initial speed.

vox = vo*cos(33)

voy = vo*sin(33)

x = 114 m

y = 10 - 1 = 9 m

distance travelle 114 m in x direction,

t = x/vox

= 114/(vo*cos(33))

now dsiplacment during the above time

y = voy*t - (1/2)*g*t^2

9 = vo*sin(33)*114/(vo*cos(33)) - (1/2)*9.8*t^2

9 = 114*tan(33) - (1/2)*9.8*t^2

==> t = 3.643 s

so, vo*cos(33) = 114/t

vo = 114/(3.643*cos(33))

= 37.3 m/s

b) t = 3.643 s

c) vx = vox

= 37.3*cos(33)

= 31.3 m/s

vy = voy - g*t

= 37.3*sin(33) - 9.8*3.643

= 15.4 m/s

speed,

v = sqrt(vx^2 +vy^2)

= sqrt(31.3^2 + 15.4^2)

= 34.9 m/s

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