What is the maximum height reached by the ball?

64

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

54

No worries! We‘ve got your back. Try BYJU‘S free classes today!

44

No worries! We‘ve got your back. Try BYJU‘S free classes today!

34

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $64$ m
Form the equation into completing the square, we get,
$h =$ $- t^{2} + 2 t + 35 + 28$
$h =$ $[- (t - 1)^{2} + 8^{2}]$
When the height is maximum, $t = 1$ , therefore, maximum height $= 64$ m

Suggest Corrections

Similar questions

A ball is thrown upwards with a velocity of $10 \frac{m}{s}$ . Find the maximum height reached by the ball.

A ball is thrown upwards with a velocity of 10 m/s. Find the maximum height reached by the ball.
Assume acceleration due to gravity $g = 10 m / s^{2}$ .

Q. A steel ball is dropped on a hard surface from a height of 1 m and rebounds to a height of 64 cm. The maximum height attained by the ball after

n^{t h}

bounce is (in m)-

Q. A ball of mass m is thrown straight upward from the top of a multi-storey building with an initial velocity of +15 m/s.
Find out the approximate maximum height reached by the ball?

Q. For an observer on trolley, direction of projection of particle is shown in the figure, while for observer on ground ball rise vertically. The maximum height reached by ball from trolley is (take

g = 10 m / s^{2}

)

Join BYJU'S Learning Program

What is the maximum height reached by the ball?

The correct option is A 64 mForm the equation into completing the square, we get,h= −t2+2t+35+28h= [−(t−1)2+82]When the height is maximum, t=1, therefore, maximum height =64 m

The correct option is A $64$ m
Form the equation into completing the square, we get,
$h =$ $- t^{2} + 2 t + 35 + 28$
$h =$ $[- (t - 1)^{2} + 8^{2}]$
When the height is maximum, $t = 1$ , therefore, maximum height $= 64$ m