Question

. A ball is launched horizontally from the edge of a table. It strikes the floor. The horizontal distance between the point on the floor directly beneath the edge of the table and the point on the floor where the ball strikes is 3.23 m. The vertical distance between the floor and the edge of the table is 1.14 m. Find (a) the time interval (s) from when the ball was launched to when the ball struck the floor and (b) the initial speed (m/s) of the ball. Next, the ball is launched from the floor at an angle 56.1 ◦ above the horizontal with the same initial speed. It again strikes the floor. Find (c) the time interval (s) from when the ball was launched to when the ball strikes the floor and (d) the horizontal distance (m) from the point where the ball was launched to the point where the ball strikes the floor.

Answer 1

answer) a) we have the equation

s=ut+1/2at²

here u=0

1.14m=1/2*9.8*t²

1.14=4.9t²

t=0.482s

answer is 0.482s

b) we have the formula

v=d/t=3.23/0.482=6.70m/s

so the answer is 6.70m/s

c) we know the formula

v²=u²+2ad

here u=0

v=6.70 m/s*tan56.1=9.97m/s

9.97²=2*9.8*d

d=5.07m

the distance of the ball above the ground=5.07-3.23=1.84m

now for time we have the formula t=( v-u)/g=9.97/9.8=1.02s

so the answer is 1.02s

d) s=1/2at²=1/2*9.8*1.02²=5.07m

so the answer 5.07m