Question

A ball is thrown at an angle of 45° to the ground. If the ball lands 79 m away, what was the initial speed of the ball? (Round the answer to the nearest whole number. g = 9.8 m/s2)

Answer 1

Answer:

The initial velocity of the ball is 28 m/s.

Explanation:

Given that,

Angle = 45°

Distance = 79 m

Using formula of horizontal and vertical distance

$s_{x}=v_{0}\cos\theta t$.....(I)

$s_{y}=v_{0}\sin\theta t-\dfrac{1}{2}gt^2$.....(II)

Put the value in both equation

$79=v_{0}\cos\theta t$

$t = \dfrac{79}{v_{0}\cos\theta}$

Now,put the value of t in equation (II)

$0=v_{0}\sin\theta(\dfrac{79}{v_{0}\cos\theta})-\dfrac{1}{2}g(\dfrac{79}{v_{0}\cos\theta})^2$

$\dfrac{1}{2}g(\dfrac{79}{v_{0}\cos\theta})^2=v_{0}\sin\theta(\dfrac{79}{v_{0}\cos\theta})$

$79\times9.8=v_{0}^2(\sin2\theta)$

$v_{0}^2=\dfrac{79\times9.8}{\sin90^{\circ}}$

$v_{0}^2=774.2\ m/s$

$v_{0}=27.82=28\ m/s\ approx$

Hence, The initial velocity of the ball is 28 m/s.